Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;
int c, a, b, n, q;
vector<double> xs;
vi sm, L, R;
double f(double x) {
  int k = lower_bound((xs).begin(), (xs).end(), x) - xs.begin();
  double re = sm[k / 2];
  if (k % 2) {
    re += x - xs[k - 1];
  }
  return re;
}
void solve() {
  cin >> c >> a >> b >> n;
  xs.resize(2 * n);
  sm.resize(2 * n + 1);
  L.resize(n);
  R.resize(n);
  for (int(i) = (0); (i) < (int)(n); ++(i)) {
    int &l = L[i], &r = R[i];
    cin >> l >> r;
    xs[2 * i] = l;
    xs[2 * i + 1] = r;
    sm[i + 1] = sm[i] + (r - l);
  }
  cin >> q;
  for (int(i) = (0); (i) < (int)(q); ++(i)) {
    int x, y;
    cin >> x >> y;
    double p = x - (double)y * (x - a) / (y - c);
    double q = x - (double)y * (x - b) / (y - c);
    double re = (f(q) - f(p)) / (q - p) * (b - a);
    cout << re << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  solve();
  return 0;
}